Answer to Question #306322 in Discrete Mathematics for kasish

Question

 Consider the recurrence relation an=4 an−1−4 an−2

Find the general solution to the recurrence relation and the solution when a0=−2 and a1=3.

Answer 1

Let us solve the recurrence relation $a_n=4 a_{n−1}−4 a_{n−2}$

The characteristic equation $k^2=4k-4$ is equivalent to $(k-2)^2=0,$

and hence has the roots $k_1=k_2=2.$

Therefore, the general solution if of the form:

$a_n=(An+B)2^n.$

Let us find the solution when $a_0=−2$ and $a_1=3.$

It follows that $-2=a_0=B$ and $3=a_1=(A+B)2=(A-2)2=2A-4.$

We conclude that $A=\frac{7}2,\ B=-2,$ and hence the solution is

$a_n=(\frac{7}2n-2)2^n.$